Problem: What is the value of the following logarithm? $\log_{5} \left(\dfrac{1}{25}\right)$
If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $5^{y} = \dfrac{1}{25}$ In this case, $5^{-2} = \dfrac{1}{25}$, so $\log_{5} \left(\dfrac{1}{25}\right) = -2$.